Code expressions in regex refer to the regex group captured, by name / relative

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

How do our code evaluations in regex refer to the regex group captured, by name and/or relative way ? Try such

'foo' =~ /(?'name'(?=o)(\w+))(?{print $name})/

# or

'foo' =~ /(?'name'(?=o)(\w+))(?{print $-1})/

Number found where operator expected at (eval 23)
[download]

to only get failure.
How is the correct Perl's way ?

Comment on Code expressions in regex refer to the regex group captured, by name / relative Download Code

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Re: Code expressions in regex refer to the regex group captured, by name / relative by haukex (Bishop) on Jan 14, 2022 at 01:49 UTC
See %+ and %-; the following prints "`oo`": `use warnings; use strict; 'foo' =~ /(?'name'(?=o)(\w+))(?{print $+{name}})/;` [download] But also note the documentation of `(?{...})`: WARNING: Using this feature safely requires that you understand its limitations. Code executed that has side effects may not perform identically from version to version due to the effect of future optimisations in the regex engine. For more information on this, see Embedded Code Execution Frequency.	[reply] [d/l] [select]
Re: Code expressions in regex refer to the regex group captured, by name / relative by LanX (Sage) on Jan 14, 2022 at 02:13 UTC
`perlretut` answers most of your questions. Previous and future ones. Update See also https://perldoc.perl.org/perlvar#Variables-related-to-regular-expressions Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery} update see Re^2: Refer by relative way to regex group (eg. the last one)	[reply]


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Code expressions in regex refer to the regex group captured, by name / relative

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